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Let $A$ be a commutative unital complex Banach algebra with norm $\|\cdot\|_A$, and let $\ell^\infty(A)$ denote all bounded sequences $(a_n)_{n\in \mathbb{N}}$ with $a_n\in A$, $n\in \mathbb{N}$, with pointwise operations and the supremum norm: $$ \|(a_n)_{n\in \mathbb{N}}\|_{\ell^\infty(A)}=\sup_{n\in \mathbb{N}} \|a_n\|_{A}. $$ Then $\ell^\infty(A)$ is itself a Banach algebra with this norm and pointwise operations. Each element in the product of the maximal ideal space of $\ell^\infty$ with the maximal ideal space of $A$ gives rise to an element of the maximal ideal space of $\ell^\infty(A)$.

Question: Can the maximal ideal space of $\ell^\infty(A)$ ever be bigger than the product of the maximal ideal space of $\ell^\infty$ with the maximal ideal space of $A$?

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    $\begingroup$ What happens if $A=\ell^\infty$? $\endgroup$
    – Matthew Daws
    Commented Jun 9, 2020 at 13:47
  • $\begingroup$ Does this show that it is strictly bigger? Can't see it yet... could you give an outline please? $\endgroup$
    – Alok
    Commented Jun 9, 2020 at 14:01
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    $\begingroup$ The maximal ideals of $l^\infty$ correspond to ultrafilters on $\mathbb{N}$. The algebra $l^\infty(l^\infty)$ is isomorphic to $l^\infty(\mathbb{N}^2)$. Is every ultrafilter on $\mathbb{N}^2$ a product of two ultrafilters on $\mathbb{N}$? $\endgroup$
    – Nik Weaver
    Commented Jun 9, 2020 at 14:54
  • $\begingroup$ Many thanks for this! $\endgroup$
    – Alok
    Commented Jun 11, 2020 at 8:22

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